Question: Let $h(x)=5\text{ln}(x)+\text{cos}(x)$. $h'(x)=$
Answer: Recall that ${\dfrac{d}{dx}[\text{ln}(x)]=\dfrac1x}$ and ${\dfrac{d}{dx}[\text{cos}(x)]=-\text{sin}(x)}$. $\begin{aligned} h'(x)&=\dfrac{d}{dx}[5\text{ln}(x)+\text{cos}(x)] \\\\ &=5{\dfrac{d}{dx}[\text{ln}(x)]}+{\dfrac{d}{dx}[\text{cos}(x)]} \\\\ &=5\cdot{\dfrac1x}+[{-\text{sin}(x)}] \\\\ &=\dfrac5x-\text{sin}(x) \end{aligned}$ In conclusion, $h'(x)=\dfrac5x-\text{sin}(x)$